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Quantum Mechanics problem

  1. Jun 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Let C|+-> = +-|+->, and consider a state |psi> = cosT|+> + sinT|->. Find T such that the product of uncertainties, deltaAdeltaB, vanishes (i.e. becomes zero).

    *Note: +- means plus or minus repectively.

    2. Relevant equations
    [A,B] = iC
    In a previous question I proved deltaAdeltaB>=1/2|<psi|C|psi>| using the Schwarz inequality and some other stuff.


    3. The attempt at a solution
    So we want 1/2|<psi|C|psi>| = 0.
    i.e. |<psi|CcosT|+> + <psi|CsinT|->| = 0.

    Then I assumed <psi| = +|cosT + <-|sinT similarly to the psi ket.

    So
    |<+|(cosT)^2.C|+> + <-|sinT.C.cosT|+> + <+|cosT.C.sinT|-> + <-|(sinT)^2.C|->| = 0

    Then use C|+-> = +-|+-> and similarly I assumed <+-|C = <+-|+- for the C bra.

    So
    |<+|(cosT)^2|+> + <-|sinTcosT|+> + <+|cosTsinT|-> + <-|(sinT)^2|->| = 0
    |(cosT)^2 + sinTcosT<-|+> + sinTcosT<+|-> + (sinT)^2| = 0
    |(cosT)^2 - sinTcosT<+|-> + sinTcosT<+|-> + (sinT)^2| = 0
    |(cosT)^2 + (sinT)^2| = 0
    which is clearly nonsense.

    Are my assumptions incorrect? Am I not allowed to convert the ket formalism to the bra formalism in this manner?
     
    Last edited: Jun 22, 2009
  2. jcsd
  3. Jun 25, 2009 #2
    I came up with a different way to do the problem using <psi| = -|cosT + <+|sinT which gives me
    |2sinTcosT| = 0
    |sin2T| = 0 by double angle formula
    T = n.Pi/2 , n an integer

    But I still don't know if what I'm doing is right (actually I suspect it's wrong).
     
  4. Jun 25, 2009 #3
    I posted this over three days ago. How long does it usually take? Surely someone can help with a second year QM problem?
     
  5. Jun 25, 2009 #4

    drizzle

    User Avatar
    Gold Member

    your work seems fine in your first post, but I doubt the signs of the products in the pre-last
    equation, try first to find out what you come up with in this:
    <epsi|epsi>=1=...
     
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